Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

When using stokes theorem to remove integrals

  1. Apr 4, 2014 #1

    joshmccraney

    User Avatar
    Gold Member

    hey pf!

    i had a question. namely, in the continuity equation we see that [tex] \frac{\partial}{\partial t}\iiint_V \rho dV = -\iint_{S} \rho \vec{v} \cdot d\vec{S}[/tex] and we may use the divergence theorem to have: [tex] \frac{\partial}{\partial t}\iiint_V \rho dV = -\iiint_{V} \nabla \cdot \big( \rho \vec{v} \big) dV[/tex]
    ultimately, we arrive at: [tex] \frac{\partial}{\partial t}\bigg( \rho \bigg) = -\nabla \cdot \big( \rho \vec{v} \big)[/tex]
    my question is, at this point, how are we able to do the following two things:
    1 interchange ##\frac{\partial}{\partial t}## inside the volume integral?
    2 drop the volume integrals entirely?

    i should say that the volume is arbitrary, and from what i remember, we have to do something like ##\lim_{V \to 0}## but i don't know the formal, mathematical procedure here.

    can someone please help me out?

    thanks!
     
  2. jcsd
  3. Apr 4, 2014 #2
    Here is some typical reasoning:
    1. The spatial coordinates themselves are independent of time, so performing time differentiation before or after has no effect on the result (provided the integration is definite).
    2. The integrals are over the same *arbitrary* volume. If they are equal then the integrands must indeed be equal. This follows naturally from the fundamental theorem of calculus for definite integration.
     
    Last edited: Apr 4, 2014
  4. Apr 4, 2014 #3

    joshmccraney

    User Avatar
    Gold Member

    but why do we need to take a limit?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: When using stokes theorem to remove integrals
  1. Stoke's theorem (Replies: 1)

  2. Stokes theorem (Replies: 1)

  3. Stokes theorem (Replies: 2)

Loading...